Answer
$x=1$
Work Step by Step
We are given the equation:
$\log_3 x+\log_3 (x+2)=1$
Use the product property of logarithms:
$\log_3 x(x+2)=1$
Write the equation in exponential form:
$x(x+2)=3^1$
$x(x+2)=3$
Solve the equation:
$x^2+2x-3=0$
$x^2+3x-x-3=0$
$x(x+3)-(x+3)=0$
$(x+3)(x-1)=0$
$x+3=0\Rightarrow x_1=-3$
$x-1=0\Rightarrow x_2=1$
Check the solutions:
$x_1=-3$
$\log_3 (-3)+\log_3 (-3+2)\stackrel{?}{=} 1$
The logarithms are not defined for $x=-3$!!!
$x_2=1$
$\log_3 1+\log_3 (1+2)\stackrel{?}{=} 1$
$0+1\stackrel{?}{=} 1$
$1=1\checkmark$
The only solution is $x=1$.
